490
Reed-Solomon code. Problem: Communicate n packets m 1 ,..., m n on noisy channel that corrupts ≤ k packets.
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.
Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,
(2) P(x) is unique degree n−1 polynomialthat contains ≥ n+k received points.
Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.
Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2· (n+2k)n−1
p0p1...
pn−1
(mod p)
Reed-Solomon code.Problem: Communicate n packets m1, . . . ,mnon noisy channel that corrupts ≤ k packets.Reed-Solomon Code:
1. Make a polynomial, P(x) of degree n−1,that encodes message: coefficients, p0, . . . ,pn−1.
2. Send P(1), . . . ,P(n+2k).
After noisy channel: Recieve values R(1), . . . ,R(n+2k).
Properties:(1) P(i) = R(i) for at least n+k points i ,(2) P(x) is unique degree n−1 polynomial
that contains ≥ n+k received points.Matrix view of encoding: modulo p.
P(1)P(2)P(3)
.
.
.P(n+2k)
=
1 1 12 · 11 2 22 · 2n−11 3 32 · 3n−1
.
.
. ·...
.
.
.1 (n+2k) (n+2k)2 · (n+2k)n−1
p0p1...
pn−1
(mod p)
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.
Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)
Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)
At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).
Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).
Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:
E(x) is error locator polynomial.Root at each error point. Degree k .
Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point.
Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.
Set up system corresponding to Q(i) = R(i)E(i) whereQ(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial.
Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1
E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal.
Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve.
Then output P(x) = Q(x)/E(x).
Berlekamp-Welsh AlgorithmP(x): degree n−1 polynomial.Send P(1), . . . ,P(n+2k)Receive R(1), . . . ,R(n+2k)At most k i ’s where P(i) 6= R(i).Idea:E(x) is error locator polynomial.
Root at each error point. Degree k .Q(x) = P(x)E(x) or degree n+k −1 polynomial.Set up system corresponding to Q(i) = R(i)E(i) where
Q(x) is degree n+k −1 polynomial. Coefficients: a0, . . . ,an+k−1E(x) is degree k polyonimal. Coefficients: b0, . . . ,bk−1,1
Matrix equations: modulo p!
1 1 · 11 2 · 2n+k−11 3 · 3n+k−1
.
.
. ·...
.
.
.1 (n+2k) · (n+2k)n+k−1
a0a1...
an+k−1
=
R(1) · 00 · 0...
.
.
. 00 · R(n+2k)
1 · 11 · 2k1 · 3k
.
.
....
.
.
.1 · (n+2k)k
b0b1...
bk−11
Solve. Then output P(x) = Q(x)/E(x).
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k
...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients.
Leading coefficient is 1.
I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.
I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1
...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient:
n+2k .
Finding Q(x) and E(x)?
I E(x) has degree k ...
E(x) = xk +bk−1xk−1 · · ·b0.
=⇒ k (unknown) coefficients. Leading coefficient is 1.I Q(x) = P(x)E(x) has degree n+k −1 ...
Q(x) = an+k−1xn+k−1 +an+k−2xn+k−2 + · · ·a0
=⇒ n+k (unknown) coefficients.
Total unknown coefficient: n+2k .
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.
an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)
...an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...
and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Solving for Q(x) and E(x)...and P(x)
For all points 1, . . . , i ,n+2k ,
Q(i) = R(i)E(i) (mod p)
Gives n+2k linear equations.an+k−1 + . . .a0 ≡ R(1)(1+bk−1 · · ·b0) (mod p)
an+k−1(2)n+k−1 + . . .a0 ≡ R(2)((2)k +bk−1(2)k−1 · · ·b0) (mod p)...
an+k−1(m)n+k−1 + . . .a0 ≡ R(m)((m)k +bk−1(m)k−1 · · ·b0) (mod p)
..and n+2k unknown coefficients of Q(x) and E(x)!
Solve for coefficients of Q(x) and E(x).
Find P(x) = Q(x)/E(x).
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0
E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0
Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)
a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)
a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example.Received R(1) = 3,R(2) = 1,R(3) = 6,R(4) = 0,R(5) = 3
Q(x) = E(x)P(x) = a3x3 +a2x2 +a1x +a0E(x) = x−b0Q(i) = R(i)E(i).
a3 +a2 +a1 +a0 ≡ 3(1−b0) (mod 7)a3 +4a2 +2a1 +a0 ≡ 1(2−b0) (mod 7)
6a3 +2a2 +3a1 +a0 ≡ 6(3−b0) (mod 7)a3 +2a2 +4a1 +a0 ≡ 0(4−b0) (mod 7)
6a3 +4a2 +5a1 +a0 ≡ 3(5−b0) (mod 7)
a3 = 1, a2 = 6, a1 = 6, a0 = 5 and b0 = 2.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.
E(x) = x−2.
1 xˆ2
+ 1 x + 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2
+ 1 x + 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2
+ 1 x + 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2
+ 1 x + 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2
----------1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2
+ 1 x + 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 5
1 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x
+ 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x
---------------x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x
+ 1
-----------------x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5
xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5
x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2
-----0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1
Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2?
1 Except at x = 2? Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2?
Hole there?
Example: finishing up.
Q(x) = x3 +6x2 +6x +5.E(x) = x−2.
1 xˆ2 + 1 x + 1-----------------
x - 2 ) xˆ3 + 6 xˆ2 + 6 x + 5xˆ3 - 2 xˆ2----------
1 xˆ2 + 6 x + 51 xˆ2 - 2 x---------------
x + 5x - 2-----
0
P(x) = x2 +x +1Message is P(1) = 3,P(2) = 0,P(3) = 6.
What is x−2x−2? 1 Except at x = 2? Hole there?
Error Correction: Berlekamp-Welsh
Message: m1, . . . ,mn.Sender:
1. Form degree n−1 polynomial P(x) where P(i) = mi .
2. Send P(1), . . . ,P(n+2k).
Receiver:
1. Receive R(1), . . . ,R(n+2k).
2. Solve n+2k equations, Q(i) = E(i)R(i) to find Q(x) = E(x)P(x)and E(x).
3. Compute P(x) = Q(x)/E(x).
4. Compute P(1), . . . ,P(n).
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor?
Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.
Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values?
Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency?
Sure. Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure.
Only n+k values.See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.
See where it is 0.
Check your undersanding.
You have error locator polynomial!
Where oh where can my bad packets be?...
Factor? Sure.Check all values? Sure.
Efficiency? Sure. Only n+k values.See where it is 0.
Hmmm...
Is there one and only one P(x) from Berlekamp-Welsh procedure?
Existence: there is a P(x) and E(x) that satisfy equations.
Hmmm...
Is there one and only one P(x) from Berlekamp-Welsh procedure?
Existence: there is a P(x) and E(x) that satisfy equations.
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)
Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:
We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1
and agree on n+2k pointsE(x) and E ′(x) have at most k zeros each.
Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.
Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.
=⇒ Q′(x)
E ′(x) =Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n
=⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Unique solution for P(x)
Uniqueness: any solution Q′(x) and E ′(x) have
Q′(x)E ′(x)
=Q(x)E(x)
= P(x). (1)Proof:We claim
Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x . (2)
Equation 2 implies 1:
Q′(x)E(x) and Q(x)E ′(x) are degree n+2k −1and agree on n+2k points
E(x) and E ′(x) have at most k zeros each.Can cross divide at n points.=⇒ Q
′(x)E ′(x) =
Q(x)E(x) equal on n points.
Both degree ≤ n =⇒ Same polynomial!
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof:
Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.
If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.
When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0.
If E ′(i) = 0, then Q′(i) = 0.=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.
When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.
When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Last bit.Fact: Q′(x)E(x) = Q(x)E ′(x) on n+2k values of x .
Proof: Construction implies that
Q(i) = R(i)E(i)
Q′(i) = R(i)E ′(i)
for i ∈ {1, . . .n+2k}.If E(i) = 0, then Q(i) = 0. If E ′(i) = 0, then Q′(i) = 0.
=⇒ Q(i)E ′(i) = Q′(i)E(i) holds when E(i) or E ′(i) are zero.When E ′(i) and E(i) are not zero
Q′(i)E ′(i)
=Q(i)E(i)
= R(i).
Cross multiplying gives equality in fact for these points.
Points to polynomials, have to deal with zeros!
Example: dealing with x−2x−2 at x = 2.
Berlekamp-Welsh algorithm decodes correctly when k errors!
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:
Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people.
Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:
n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).
Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses:
Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).
Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.
Only one polynomial contains n+kEfficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.
Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!
Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:
Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.
Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.
Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..
allows for efficiency. Magic of polynomials.Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency.
Magic of polynomials.Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Summary: polynomials.Set of d +1 points determines degree d polynomial.
Encode secret using degree k −1 polynomial:Can share with n people. Any k can recover!
Encode message using degree n−1 polynomail:n packets of information.
Send n+k packets (point values).Can recover from k losses: Still have n points!
Send n+2k packets (point values).Can recover from k corruptionss.Only one polynomial contains n+k
Efficiency.Magic!!!!Error Locator Polynomial.
Relations:Linear code.Almost any coding matrix works.Vandermonde matrix (the one for Reed-Solomon)..allows for efficiency. Magic of polynomials.
Other Algebraic-Geometric codes.
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative,
inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!
Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ?
Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)?
Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...
maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)?
Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh..
−3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).
Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.
But can you get closer? Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. −3 = 4 (mod 7).Another problem.
4 is close to 3.But can you get closer?
Sure. 3.5. Closer. Sure? 3.25, 3.1,3.000001. . . .
For reals numbers we have the notion of limit, continuity, andderivative.......
....and Calculus.
For modular arithmetic...no Calculus. Sad face!
Farewell to modular arithmetic...Modular arithmetic modulo a prime.
Add, subtract, commutative, associative, inverses!Allow for solving linear systems, discussing polynomials...
Why not modular arithmetic all the time?
4 > 3 ? Yes!
4 > 3 (mod 7)? Yes...maybe?
−3 > 3 (mod 7)? Uh oh.. �
